Question

A \(4.00 \cdot 10^{3}-n F\) parallel plate capacitor is connected to a \(12.0-\mathrm{V}\) battery and charged. a) What is the charge \(Q\) on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor? The \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor is then disconnected from the \(12.0-\mathrm{V}\) battery and used to charge three uncharged capacitors, a \(100 .-n F\) capacitor, a \(200 .-\mathrm{nF}\) capacitor, and a \(300 .-\mathrm{nF}\) capacitor, connected in series. c) After charging, what is the potential difference across each of the four capacitors? d) How much of the electrical energy stored in the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor was transferred to the other three capacitors?

Short answer

Also, determine the potential difference across each of the four capacitors and the amount of electrical energy transferred to the other three capacitors. a) The charge on the positive plate is \(4.8 \cdot 10^{-5} \text{ C}\). b) The electric potential energy stored in the capacitor is \(2.88 \cdot 10^{-4} \text{ J}\). c) The potential difference across each of the four capacitors is: the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor: 12.0 V, the \(100\ \mathrm{nF}\) capacitor: 480.0 V, the \(200 \text{ nF}\) capacitor: 240.0 V, and the \(300 \text{ nF}\) capacitor: 160.0 V. d) The amount of electrical energy transferred from the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor to the other three capacitors is \(5.6 \cdot 10^{-5} \text{ J}\).

Step-by-step solution

Calculate the charge on the positive plate of the capacitor

Using the formula to calculate the charge on a capacitor: \(Q = C \times V\), where \(Q\) is the charge, \(C\) is the capacitance, and \(V\) is the voltage. We have \(C = 4.00 \cdot 10^{3} \text{ nF} = 4.0 \cdot 10^{-6} \text{ F}\) and \(V = 12.0 \text{ V}\). So, \(Q = (4.0 \cdot 10^{-6} \text{ F})(12.0 \text{ V}) = 4.8 \cdot 10^{-5} \text{ C}\). #a) The charge \(Q\) on the positive plate of the capacitor is \(4.8 \cdot 10^{-5} \text{ C}\).

Calculate the electric potential energy stored in the capacitor

Using the formula for electric potential energy: \(U = \frac{1}{2}CV^2\), where \(U\) is the electric potential energy, \(C\) is the capacitance, and \(V\) is the voltage. So, \(U = \frac{1}{2}(4.0 \cdot 10^{-6} \text{ F})(12.0 \text{ V})^2 = 2.88 \cdot 10^{-4} \text{ J}\). #b) The electric potential energy stored in the capacitor is \(2.88 \cdot 10^{-4} \text{ J}\).

Calculate the total capacitance of the three capacitors in series

Using the formula for the total capacitance of capacitors in series: \(\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\). We have \(C_1 = 100\ \mathrm{nF} = 1.0 \cdot 10^{-7}\ \mathrm{F},\ C_2 = 200\ \mathrm{nF} = 2.0 \cdot 10^{-7}\ \mathrm{F}\), and \(C_3 = 300\ \mathrm{nF} = 3.0 \cdot 10^{-7}\ \mathrm{F}\). \(\frac{1}{C_T} = \frac{1}{1.0 \cdot 10^{-7}\ \mathrm{F}} + \frac{1}{2.0 \cdot 10^{-7}\ \mathrm{F}} + \frac{1}{3.0 \cdot 10^{-7}\ \mathrm{F}}\) To find \(C_T\), take the reciprocal of the sum: \(C_T = \frac{1}{(\frac{1}{1.0 \cdot 10^{-7}\ \mathrm{F}} + \frac{1}{2.0 \cdot 10^{-7}\ \mathrm{F}} + \frac{1}{3.0 \cdot 10^{-7}\ \mathrm{F}})} = 6.00 \cdot 10^{-8} \text{ F}\)

Calculate the potential difference across each capacitor in series

When the capacitors are connected in series, the charge on each capacitor will be equal. We have already calculated the charge on the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor, which is \(Q = 4.8 \cdot 10^{-5} \text{ C}\). To calculate the potential difference across each capacitor, use the formula \(V_i = \frac{Q}{C_i}\), where \(V_i\) is the potential difference across the \(i\)th capacitor and \(C_i\) is the capacitance of the \(i\)th capacitor. For the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor: \(V_1 = \frac{4.8 \cdot 10^{-5} \text{ C}}{4.0 \cdot 10^{-6} \text{ F}} = 12.0 \text{ V}\) For the \(100\ \mathrm{nF}\) capacitor: \(V_2 = \frac{4.8 \cdot 10^{-5} \text{ C}}{1.0 \cdot 10^{-7} \text{ F}} = 480.0 \text{ V}\) For the \(200 \text{ nF}\) capacitor: \(V_3 = \frac{4.8 \cdot 10^{-5} \text{ C}}{2.0 \cdot 10^{-7} \text{ F}} = 240.0 \text{ V}\) For the \(300 \text{ nF}\) capacitor: \(V_4 = \frac{4.8 \cdot 10^{-5} \text{ C}}{3.0 \cdot 10^{-7} \text{ F}} = 160.0 \text{ V}\) #c) The potential difference across each of the four capacitors are: the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor: 12.0 V, the \(100\ \mathrm{nF}\) capacitor: 480.0 V, the \(200 \text{ nF}\) capacitor: 240.0 V, and the \(300 \text{ nF}\) capacitor: 160.0 V.

Calculate the energy transferred to the other three capacitors

To calculate the total electric potential energy stored in the three capacitors connected in series, use the formula \(U = \frac{1}{2} C_TV_T^2\), where \(U\) is the electric potential energy, \(C_T\) is the total capacitance, and \(V_T\) is the total potential difference across the series capacitance. The total potential difference across the series capacitance is the sum of the potential differences across each capacitor: \(V_T = V_2 + V_3 + V_4 = 480.0 \text{ V} + 240.0 \text{ V} + 160.0 \text{ V} = 880.0 \text{ V}\). So, \(U_T = \frac{1}{2} (6.00 \cdot 10^{-8} \text{ F})(880.0 \text{ V})^2 = 2.32 \cdot 10^{-4} \text{ J}\). The energy transferred to the other three capacitors is the difference between the initial energy stored in the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor and the energy stored in the other three capacitors: \(\Delta U = 2.88 \cdot 10^{-4} \text{ J} - 2.32 \cdot 10^{-4} \text{ J} = 5.6 \cdot 10^{-5} \text{ J}\). #d) The amount of electrical energy transferred from the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor to the other three capacitors is \(5.6 \cdot 10^{-5} \text{ J}\).

Key concepts

Electric Potential Energy

Electric potential energy is a key concept when learning about capacitors in the context of electrical circuits. It represents the energy a charged object possesses due to its position in an electric field. In the case of capacitors, this energy is stored between the plates as a result of the electrical separation of charges.

For a capacitor, the formula for electric potential energy (U) is given by \( U = \frac{1}{2}CV^2 \), where C is the capacitance and V is the potential difference across the capacitor. When a capacitor is charged by a battery as in our exercise, the energy is stored and can do work later on, for instance, when the capacitor is discharged. Understanding how potential energy is calculated and stored within a capacitor is essential for analyzing electrical circuits and their energy conversion efficiency.

Series Capacitors

Capacitors connected in series present a fascinating topic in the study of electric circuits. When we connect capacitors in series, the total capacitance is not simply the sum of each capacitor's capacitance; rather, it is determined by the reciprocal of the sum of reciprocals of the individual capacitances.

This results because the charge on each capacitor in series must be the same. To find the total capacitance (\(C_T\)) of a series circuit, we use the formula \( \frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n} \). After finding the reciprocal of the sum, you’ll get the equivalent capacitance of the series configuration. In our exercise, this principle was crucial in understanding how the charge and voltage distributed across the capacitors once they are connected.

Potential Difference

Potential difference is another fundamental concept in the realm of electronics. It represents the difference in electric potential between two points in an electric field. When dealing with capacitors, the potential difference (often referred to as voltage) across a capacitor dictates how much charge it can store.

In a series circuit, like the one presented in the exercise, the potential difference across each capacitor can be determined once we know the charge on them. Using the formula \( V = \frac{Q}{C} \), where Q is the charge and C is the capacitance, we can calculate the voltage across each capacitor. For students, it's crucial to grasp that even though the charge remains consistent across capacitors in series, the potential differences may vary according to their individual capacitance.

Stored Electrical Energy

Stored electrical energy within a capacitor is what makes capacitors such useful components in various electronic devices. This energy is a direct result of the electric field created between the plates of a capacitor when it is charged.

As demonstrated in our exercise, calculating the energy stored can be done using the formula \( U = \frac{1}{2} C V^2 \), which shows us that the energy stored is proportional to the square of the voltage across the capacitor and directly proportional to its capacitance. When capacitors are used in series, their stored energy is not simply additive. Instead, it must be calculated based on the series equivalent capacitance and the total potential difference across the series. Understanding how this energy is stored and later dispersed is vital for any student studying electric circuits or energy management in electronic systems.